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QUESTION: 1

The domain of the function is

Solution:

For f to be defined term under square root has to be greater than zero.

⇒ cosx − 1 ≥ 0

⇒ cosx ≥ 1

Only possible if cosx = 1

⇒ x = 2nπ ∀ n∈Z

QUESTION: 2

The domain of the function

Solution:

Since you want the square root, 1-cos(x) must not be negative.

Solve 1-cos(x) ≥ 0

cos(x) ≤ 1

But -1 ≤ cos(x) ≤ 1 for all "x".

So Domain is All Real Numbers.

QUESTION: 3

The domain of the function

Solution:

QUESTION: 4

The domain of the function

Solution:

QUESTION: 5

Let x be any real, then [x + y] = [x] + [y] holds for

Solution:

Let y has some fractional part

x = 6.5

y = 0.9

[x] = 6, [y] = 0, [x]+[y] = 6

[x + y] = [7.4] = 7

[x + y] ≠ [x]+[y]

Let x, y be integers

x = 6.5

y = 1

[x] = 6, [y] = 1, [x]+[y] = 7

[x + y] = [7.5] = 7

[x + y] = [x]+[y]

So, [x + y] = [x]+[y] holds for y ϵ I

QUESTION: 6

The function f (x) = x + cos x

Solution:

f(x) = x

g(x) = cosx

D(f) = R

D(g) = R

D(f+g) = D(f) ⋂ D(g)

So , D(f+g) = R

Therefore, given function is defined for real numbers.

QUESTION: 7

Solution:

lim(x → π) sinx/x-π

lim(h → 0) sin(π-h)/π-h-π

lim(h → 0) -sinh/h

⇒ -1

QUESTION: 8

A function / from the set of natural numbers to integers defined by

Solution:

Clearly/is both one - one and onto Because if n is odd, values are set of all non-negative integers and if n is an even, values are set of all negative integers.

QUESTION: 9

If f (x) = and g (x) = then domain of fog is

Solution:

QUESTION: 10

If then domain of fof is

Solution:

QUESTION: 11

If then

Solution:

QUESTION: 12

The interval in which the values of f (x) lie where f (x) = 3 sin

Solution:

f(x) = 3.sin(√π^{2}/16 − x^{2})

since the quantity within the square root can not be negative.

therefore

π^{2}/16 − x^{2} ≥ 0

x^{2} ≤ π^{2}/16

−π/4 ≤ x ≤ π/4

the minimum value of the function is 3.sin0=0

and the maximum value of the function is 3sinπ/4 = 3/√2

therefore range is [0, 3/√2]

QUESTION: 13

The range of the function

Solution:

y = (x^{2} - 3x + 2)/(x^{2} + x -6)

= [(x-2)(x-1)]/[(x-2)(x+3)]

= (x-1)/(x+3)

= 1 - 4/(x+3)

x is not equal to 2 and -3

For x = 2, y = 1/5

for x = -3, y → 1

Range (y): R - {1, 1/5}

QUESTION: 14

If f(x) = is equal to

Solution:

f(x) = log[(1+x)/(1-x)] ……….(1)

Substitute 2x/1+x^{2} in place of x in equation(1)

f(2x1+x^{2})=log(1+2x1+x^{2})(1−2x1+x^{2})

⟹f(2x/(1+x^{2}))=log[1+(2x/(1+x^{2}))/(1-(2x/1+x^{2}))]

⟹f(2x/(1+x^{2}))=log((1+x^{2}+2x)/(1+x^{2}−2x))

⟹f(2x/(1+x^{2}))=log((1+x)^{2} / (1−x))^{2}

⟹f(2x/ (1+x^{2}))=log(1+x / 1−x)^{2}

We know that logm^{n}=nlogm

⟹f(2x/(1+x^{2}))=2log(1+x / 1−x)

From equation (1):

f(2x/(1+x^{2}))=2f(x)

QUESTION: 15

If is equal to

Solution:

QUESTION: 16

Which of the following functions is not one-one ?

Solution:

QUESTION: 17

If f (x) then f (1) is equal to

Solution:

QUESTION: 18

The domain of the real-valued function

Solution:

f(x) = (x-3)(x-1)/[(x2 -4)^1/2]

= x2 - 4 = 0

(x-2)(x+2) = 0

x = 2 and -2

(-∞, -2 ) U (2,∞)

QUESTION: 19

A condition for a function y = f (x) to have an inverse is that it should be

Solution:

For a function to have its inverse in a given domain, it should be continuous in that domain and should be a one-one function in that domain.

If the function is one-one in the domain, then it has to be strictly monotonic.

For example y=sin(x) has its domain in xϵ[−π/2,π/2] since it is strictly monotonic and continuous in that domain.

QUESTION: 20

Set A has 3 elements and set B has 4 elements. The number of injections that can be defined from A to B is

Solution:

Given that n(A)=3 and n(B)=4, the number of injections or one-one mapping is given by.

QUESTION: 21

The range of the function

Solution:

yx^{2} + xy + y = x^{2} - x + 1

(y-1)x^{2} + (y+1)x + y-1 = 0

if y = -1

-2x^{2} -1 -1 = 0

-2x^{2} -2 = 0

Therefore, y cannot be equal to -1 since x will be a complex value

Now, y not equal to -1.

(y-1)x^{2} + (y+1)x + y-1 = 0 has real roots

hence, (y+1)^{2} - 4 (y-1)(y-1) ≥ 0

y2 + 2y + 1 - 4y^{2} + 8y - 4 ≥ 0

-3y^{2} + 10y - 3 ≥ 0

3y^{2} - 10y + 3 ≤ 0

3y^{2} - 9y - y + 3 ≤ 0

3y (y-3) - 1 (y-3) ≤0

(3y-1)(y-3) ≤ 0

y ∈ [1/3,3]

Hence, range of function is [1/3,3]

QUESTION: 22

On the set Z of all integers define f ; Z → Z as follows : f (x) = x/2 if x is even, and f (x) = 0 if x is odd , then f is

Solution:

f(x) = {x/2 x is even, 0 x is odd}

f(1) = f(3) = f(5) = f(odd) = 0

Hence f is not one - one.

column = Z

Range = {0, even}

= {0,2,4,6,8,10.......}

f(x) = x/2 implies even

f(odd) = 0

Range is not equal to codomain

Hence, onto

QUESTION: 23

Let f : R → R be defined by f (x) = 3x – 4, then f^{-1} (x) is equal to

Solution:

f(x) = 3x - 4

Let f(x) = y

y = 3x - 4

y + 4 = 3x

x = (y+4)/3

x → f^{-1}(x) and y → x

f-1(x) = (x+4)/3

QUESTION: 24

The number of bijective functions from the set A to itself when A constrains 106 elements is

Solution:

The number of bijections from A to A, will be n! for n elements is A.

Now there are 106, elements.

Hence the number of bijections will be n!=(106)!

QUESTION: 25

In the set W of whole numbers an equivalence relation R defined as follow : aRb iff both a and b leave same remainder when divided by 5. The equivalence class of 1 is given by

Solution:

aRb iff both a and b leave same remainder when divided by 5.

a= 5x+n

a-1= 5x

b=5y+n

b-1 =5y

Equivalence class = {1,6,11,16,21}

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